#1

ok guys i need your help. there is this math word problem i am having trouble with..

A farmer is taking her eggs to the market in a cart, but she hits a

pothole, which knocks over all the containers of eggs. Though she is

unhurt, every egg is broken. So she goes to her insurance agent, who

asks her how many eggs she had. She says she doesn't know, but she

remembers somethings from various ways she tried packing the eggs.

When she put the eggs in groups of two, three, four, five, and six

there was one egg left over, but when she put them in groups of seven

they ended up in complete groups with no eggs left over.

What can the farmer figure from this information about the number of

eggs she had? Is there more than one answer?

now, obviously the lowest possible answer is 301. but i am supposed to find an equation to help determine any other amount of eggs the farmer can have, and i have no idea how to do that. any help would be greatly appreciated.

thanks in advance guys

A farmer is taking her eggs to the market in a cart, but she hits a

pothole, which knocks over all the containers of eggs. Though she is

unhurt, every egg is broken. So she goes to her insurance agent, who

asks her how many eggs she had. She says she doesn't know, but she

remembers somethings from various ways she tried packing the eggs.

When she put the eggs in groups of two, three, four, five, and six

there was one egg left over, but when she put them in groups of seven

they ended up in complete groups with no eggs left over.

What can the farmer figure from this information about the number of

eggs she had? Is there more than one answer?

now, obviously the lowest possible answer is 301. but i am supposed to find an equation to help determine any other amount of eggs the farmer can have, and i have no idea how to do that. any help would be greatly appreciated.

thanks in advance guys

#2

Over 9000

amidoinitrite?

amidoinitrite?

#3

What about 49?

#4

7x=y

2x=y + 1

3x=y+1

etc

edit i didn't think this one out too much so this is probably wrong but worth a shot.

2x=y + 1

3x=y+1

etc

edit i didn't think this one out too much so this is probably wrong but worth a shot.

#5

49 doesnt work for 5

#6

7x=y

2x=y + 1

3x=y+1

etc

edit i didn't think this one out too much so this is probably wrong but worth a shot.

This, your right here.

Then use simultanious equations to i think.

#7

least common multiplier plus 1 (2x3x7 plus 1)

301

= 2 X 150 + 1

= 3 X 100 + 1

= 4 X 75 + 1

= 5 X 60 + 1

= 6 X 50 + 1

= 7 X 43

301

= 2 X 150 + 1

= 3 X 100 + 1

= 4 X 75 + 1

= 5 X 60 + 1

= 6 X 50 + 1

= 7 X 43

#8

so glad I'm out of school!

#9

There are multiple answers. Any prime number (other than 2,3, and 5) times seven will yield an answer that works. This is because the factors won't include the numbers 2,3,4,5,6. Therefore, other than 7, you can do 49, 77, 91, 119, and so on.

Hope that helps dude..!

Hope that helps dude..!

#10

ok here we go the correct solution. x is the number of eggs she has so when x is divided by 7 it should equal one. now when x is divided by say 2 for two groups if there is one egg remaining then x/2 +1/2 should equal 1. i'm not 100 percent on the second part but i know the first part is true

#11

easiest way is to take 7 and keep multiplying it until you find all the values of it that end up with a 6 or 1 in the ones place. From then you just have to -1 from those and divide it by the other numbers until they all come up with 1s.

#12

m<A + m<B = eleventy billion

#13

x=*7

x=*2+1

x=*3+1

x=*4+1

x=*5+1

x=*6+1

The number is multiple of 2,3,5 plus 1, and multille of 7

IF the number is 13, it is 12+1, so it is *6+1, *2+1, and *3+1 at once..

If the number is 5, it is 4+1, so it is both *2+1, and *4+1

You get then

x=*7

x=*4+1

x=*6+1

x=*5+1

You want to have an ecuation with a variable n, such as as when n equals something, the result equals x and x has all those properties...

I think you should do some kind of inductive reasoning here, but get the equation first (which I don't know now)

x=*2+1

x=*3+1

x=*4+1

x=*5+1

x=*6+1

The number is multiple of 2,3,5 plus 1, and multille of 7

IF the number is 13, it is 12+1, so it is *6+1, *2+1, and *3+1 at once..

If the number is 5, it is 4+1, so it is both *2+1, and *4+1

You get then

x=*7

x=*4+1

x=*6+1

x=*5+1

You want to have an ecuation with a variable n, such as as when n equals something, the result equals x and x has all those properties...

I think you should do some kind of inductive reasoning here, but get the equation first (which I don't know now)